The first step is to determine the domain of the functions involved. With the restriction #0<=x<=360# (degrees, I assume) we have exclude the points where #tan(x)# is not defined.
By definition, #tan(x)=sin(x)/cos(x)#
Therefore, we have to exclude points where #cos(x)=0#, that is the domain is described as
#x != 90# and #x != 270#
With the above restrictions in mind we can transform the equation as follows:
#3sin(x)+2sin(x)/cos(x) = 0#
The immediate temptation to reduce the above equation by #sin(x)# should be accompanied by checking if it can be equal to zero to avoid losing solutions.
Indeed, #sin(x)=0# can occur within a domain defined above.
It happens when
#x=0#, #x=180# and #x=360#.
So the three values above are solutions.
After specifying these three solutions we can reduce the equation by #sin(x)# getting
#3+2/cos(x)=0#,
from which follows
#cos(x)=-2/3#,
solutions of this are
#x = arccos(-2/3)# and #x = -arccos(-2/3)#
CHECKING
The first three solutions (#0#, #180# and #360#) cause both #sin(x)# and #tan(x)# to be equal to #0#, therefore the left side will be equal to #0#.
The next two values (#arccos(-2/3)# and #-arccos(-2/3)# result in the following:
#3sin(arccos(-2/3))+2sin(arccos(-2/3))/cos(arccos(-2/3)) =#
#= sin(arccos(-2/3))*[3+2/(-2/3)] = #
#=sin(arccos(-2/3))*[3-3] = 0#
#3sin(-arccos(-2/3))+2sin(-arccos(-2/3))/cos(-arccos(-2/3)) =#
#= sin(-arccos(-2/3))*[3+2/(-2/3)] = #
#=sin(-arccos(-2/3))*[3-3] = 0#
All is checked.
